题目

Problem

Figure 1 shows a sketch of part of the curve with equation

y=\frac{9}{(2x-3)^{1.25}}\qquad x>\frac{3}{2}

The finite region $R$ , shown shaded in Figure 1, is bounded by the curve, the line with equation $y=9$ and the line with equation $x=6$

This region is rotated through $2\pi$ radians about the $x$ -axis to form a solid of revolution.

Find, by algebraic integration, the exact volume of the solid generated.

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题目中文翻译

图 1 给出了曲线一部分的草图，其方程为

y=\frac{9}{(2x-3)^{1.25}}\qquad x>\frac{3}{2}

图中阴影部分所示的有限区域 $R$ 由该曲线、方程为 $y=9$ 的直线以及方程为 $x=6$ 的直线围成。

将该区域绕 $x$ 轴旋转 $2\pi$ 弧度，形成一个旋转体。

求该旋转体的精确体积，使用代数积分。

解答

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先求曲线与 $y=9$ 的交点，确定区域的左端点。旋转后，直线 $y=9$ 形成外圆柱，而曲线下方的区域形成需要扣除的内层旋转体；分别求出两者体积后相减。

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At the left-hand boundary of $R$ ,

\frac{9}{(2x-3)^{1.25}}=9.

Since $x>\frac32$ , we have $2x-3>0$ . Therefore,

(2x-3)^{1.25}=1,

which gives $x=2$ .

The volume generated by rotating the region under the curve from $x=2$ to $x=6$ is

\begin{align*} V_{\text{inner}} =&\,\pi\int_2^6 y^2\,\mathrm{d}x\\ =&\,81\pi\int_2^6 (2x-3)^{-5/2}\,\mathrm{d}x\\ =&\,\pi\bigg[ -27(2x-3)^{-3/2} \bigg]_2^6\\ =&\,\pi(-1+27)\\ =&\,26\pi. \end{align*}

The line $y=9$ generates a cylinder of radius $9$ and length $6-2=4$ , whose volume is

\begin{align*} V_{\text{outer}} =&\,\pi(9^2)(4)\\ =&\,324\pi. \end{align*}

Hence the required volume is

\begin{align*} V =&\,V_{\text{outer}}-V_{\text{inner}}\\ =&\,324\pi-26\pi\\ =&\,\boxed{298\pi}. \end{align*}